在C中迭代相同类型的结构成员 [英] Iterating over same type struct members in C

问题描述

是否可以使用指针迭代所有成员都属于同一类型的 C 结构.以下是一些无法编译的示例代码:

Is it possible to iterate of a C struct, where all members are of same type, using a pointer. Here's some sample code that does not compile:

#include <stdio.h>
#include <stdlib.h>

typedef struct
{
    int mem1 ;
    int mem2 ;
    int mem3 ;
    int mem4 ;
} foo ;

void my_func( foo* data )
{
    int i ;
    int* tmp = data ; // This line is the problem

    for( i = 0; i < 4; ++i )
    {
        ++tmp ;
        printf( "%d
", *tmp ) ;
    }
}

int main()
{
    foo my_foo ;
    //
    my_foo.mem1 = 0 ;
    my_foo.mem2 = 1 ;
    my_foo.mem3 = 2 ;
    my_foo.mem4 = 3 ;
    //
    my_func( &my_foo ) ;
    return 0 ;
}

假设您的编译器/内核不会尝试为缓冲区溢出提供堆栈保护，则 foo 的成员应该在内存中一个接一个地对齐.

The members of foo should be aligned in memory to be one after another, assuming your compiler/kernel does not try to provide stack protection for buffer overflow.

所以我的问题是:

我将如何迭代 C 结构中具有相同类型的成员.

How would I iterate over members of a C struct that are of the same type.

推荐答案

最简单的方法是创建一个联合，一部分单独包含每个成员，一部分包含数组.我不确定依赖于平台的填充是否会干扰对齐.

The easiest way would be to create a union, one part which contains each member individually and one part which contains an array. I'm not sure if platform-dependent padding might interfere with the alignment.

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